An intro to real analysis question. eazy?

eibon
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Homework Statement



Let f : A -> B be a bijection. Show that if a function g is such that f(g(x)) = x for
all x ϵ B and g(f(x)) = x for all x ϵ A, then g = f^-1. Use only the definition of a
function and the definition of the inverse of a function.


Homework Equations





The Attempt at a Solution


well since f is a bijection then there exist an f^-1 that remaps f back to A and since g does that then g is the unique inverse of f, or something like that please help I am not very good that this stuff
 
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Yes, you only need to prove uniqueness of an inverse function. If f(g(x))=x, then g(x) must be the unique argument a for which f(a) = x. And so, this condition completely defines g for all the arguments. g is then obviously equal to f-1.
 
thanks losiu for your response.
but how do you prove that it is the unique inverse?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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